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| Section | Objectives |
|---|---|
| Statistical Inference | - Hypothesis Testing (Introductory Level)
|
| Regression and Modeling | - Linear Relationships
|
| Descriptive Statistics | - Two Variable Data Analysis
|
| Probability Theory | - Fundamental Probability Concepts
|
>> New Applied-Probability-and-Statistics Exam Vce <<
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NEW QUESTION # 129
A fitness center owner notices that gym attendance increases as the number of daylight hours increases. The owner calculates a correlation coefficient between daylight hours and gym attendance of r = 0.72.
Based on this information, what can be concluded?
Answer: B
Explanation:
The correlation coefficient r measures the direction and strength of a linear association between two quantitative variables. Since r = 0.72 is positive, the relationship is positive: as daylight hours increase, gym attendance tends to increase. The value 0.72 also indicates a moderately strong to strong linear association because it is closer to 1 than to 0. However, correlation alone does not establish causation. Even though daylight hours and gym attendance move together, other variables could influence attendance, such as weather, seasonal routines, work schedules, or fitness promotions. Therefore, the valid conclusion is that there is a positive association, not a proven causal relationship. Options A and B are incorrect because they describe a negative relationship, which conflicts with the positive correlation coefficient. Option D overstates the evidence by claiming causation. References/topics from the Study Guide: correlation coefficient, positive association, linear relationship, correlation versus causation.
NEW QUESTION # 130
Expected value of X = 1×0.2 + 2×0.5 + 3×0.3 = ?
Answer: B
Explanation:
Expected value is the long-run average value of a random variable. For a discrete random variable, it is calculated by multiplying each possible value by its probability and then adding those products. Here, the expression is already structured as an expected value calculation: 1×0.2 + 2×0.5 + 3×0.3. Compute each product: 1×0.2 = 0.2, 2×0.5 = 1.0, and 3×0.3 = 0.9. Add them: 0.2 + 1.0 + 0.9 = 2.1. Therefore, the expected value is 2.1. This does not mean the random variable must equal 2.1 in a single trial; it means that over many repetitions, the average outcome would approach 2.1. Option B is close but omits part of the weighted contribution. Options C and D do not match the weighted-average computation. Study Guide references
/topics: expected value, discrete random variables, weighted average, probability distributions.
NEW QUESTION # 131
Determine z in the equation:
1/(6z) = 2/9
Answer: B
Explanation:
The equation 1/(6z) = 2/9 is a proportion because it sets two fractions equal to each other. To solve it, use cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction, and multiply the denominator of the first fraction by the numerator of the second fraction: 1 × 9 = 2 × 6z. This gives 9 = 12z. To isolate z, divide both sides by 12: z = 9/12. This fraction can also be simplified to 3/4, but the answer choices list the unsimplified equivalent form 9/12. The other options do not satisfy the original equation. For example, if z = 2/54, then 6z is very small and the left side becomes much larger than 2/9. If z =
12/9 or 18/2, the denominator becomes too large, making the left side too small. References/topics from the Study Guide: proportions, solving equations, cross multiplication, equivalent fractions.
NEW QUESTION # 132
Binomial expected value = ?
Answer: B
Explanation:
For a binomial random variable, the expected value is E(X) = np, where n is the number of trials and p is the probability of success on each trial. The binomial model applies when there is a fixed number of independent trials, each trial has two outcomes, and the probability of success remains constant. The expected value represents the long-run average number of successes across repeated binomial experiments. For example, if a test has 20 independent questions and the probability of success on each is 0.75, the expected number correct is 20 × 0.75 = 15. Option B, n/p, and option C, p/n, do not represent the binomial mean. Option D, n + p, incorrectly adds the number of trials and success probability, which are not comparable units. The formula np captures both how many opportunities there are and how likely success is on each opportunity. Study Guide references/topics: binomial distribution, expected value, success probability, discrete random variables.
NEW QUESTION # 133
Standard error decreases when:
Answer: C
Explanation:
The standard error of the sample mean is calculated as SE = s/#n, where s is the sample standard deviation and n is the sample size. Since n appears in the denominator, increasing the sample size reduces the standard error, assuming the standard deviation stays constant. This reflects the fact that larger samples produce more stable estimates of the population mean. Option B is incorrect because decreasing sample size increases standard error. Option C is incorrect because a larger standard deviation increases standard error by increasing the numerator. Option D is incorrect because the mean affects the center of the distribution but does not directly determine the standard error. Standard error is about precision: smaller standard error means less sampling variability and a more precise estimate. That is why larger samples are preferred in inference. Study Guide references/topics: standard error, sample size, standard deviation, sampling variability.
NEW QUESTION # 134
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